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Energía electrostática
Capacitores
Campo eléctrico en materiales
Capacitores
Capacitores
Un capacitor es un dispositivo que almacena carga eléctrica. Está formado por
dos conductores próximos uno a otro, separados por un aislante, de tal modo que
puedan estar cargados con el mismo valor, pero con signos contrarios.
Un capacitor entonces acumula una carga –Q en un conductor
y una carga +Q en el otro conductor
La carga acumulada crea un campo
+Q
–Q
eléctrico y por lo tanto, una diferencia de
potential entre dichos conductores.
Capacitancia
La capacitancia de un dispositivo se define como
Q
C=
V
Q es la carga acumulada en un conductor
V es la diferencia de potential entre los
conductores
La unidad es el faradio (F) = Culombio/Voltio
(Remember: el culombio y por ende el faradio es una unidad enorme, de modo que en
dispositivos ‘usuales’ las unidades que encontraremos son multiplos pequeños
μF = 10-6 F o bien pF = 10-12 F
4
Capacitancia de un
Conductor esférico
El potencial en la superficie de un conductor
esférico cargado con una carga Q y de radio R es
Q
V =k
R
De modo que su capacitancia es
Q R
C= = = 4πε 0 R
V k
5
Capacitancia de un capacitor de placas
paralelas
Cuando dos placas conductoras de area A
están separadas por una distancia (pequeña
comparada con las dimensiones de las
placas) d el campo eláctrico entre ellas será
aproximadamente constante y de módulo
=
E σ=
/ ε 0 Q /(ε 0 A)
6
Capacitancia de un capacitor de placas
paralelas
Como
el
campo
eléctrico
es
constante, la diferencia de potencial
V = Ed
/ ε 0 )d Qd /(ε 0 A)
= (σ
=
entre las placas es
entonces la capacitancia es
C=
7
ε0 A
d
Capacitance
Cylindrical Capacitors
A coaxial cable of length L is an example
of a cylindrical
capacitor
2πε 0 L
C=
ln( R2 / R1 )
R2
R1
8
Energía Electrostática
Electrostatic Energy
Work is required to assemble a charge distribution
a
q2
q1
kq1
=
W2 q=
q2
2V1
a
a
a
q3
Total work done
kq1 kq2
W3 = q3 (V1 + V2 )= q3 (
+
)
a
a
kq1q2 kq1q3 kq2 q3
W = W2 + W3 =
+
+
a
a
a
Electrostatic Energy
The work dW required to add an element of charge dq
to an existing charge distribution is
dW = dqV
dq
where V is the potential at the final
location of the charge element. The
total work required is therefore
Since the electric is
=
W =
dW
Vdq conservative, the work is
stored as electrostatic energy,
U.
∫
∫
Storage of Electrostatic Energy
Work must be done to move positive charge
from a negatively charged conductor to one
that is positively charged. Or to move
negative charge in the reverse direction.
Storage of Electrostatic Energy
In moving charge dq, the electrostatic
energy of the capacitor is increased by dU = Vdq
2
q
1Q
Therefore,
=
U ∫=
dq
0 C
2 C
1
= QV
2
1
2
U = CV
2
Q
Energy Density of Electric Field
E = Q /(ε 0 A)
V = Ed
Electric field
Electric potential
1
U = QV
2
Potential energy
1
1
2
U =
(ε 0 EA)( Ed )
=
ε 0 E ( Ad )
2
2
14
Energy Density of Electric Field
The energy density uE
energy
=
uE = U /( Ad )
volume
1
2
uE = ε 0 E
2
15
This expression
holds true for
any electric field
Example – A Thunderstorm
How much electrical energy is stored in a
typical thundercloud?
Assume a cloud of
height h = 10 km,
radius r = 10 km,
with a uniform
electric field
E = 105 V/m.
16
http://redcrossggr.files.wordpress.com/2008/06/thunderstorm.jpg
Example – A Thunderstorm
Narrative
The problem is about stored electric energy.
Since the electric field, E, is uniform, so to is
the energy density uE = ½ ε0 E2 in the cloud.
Therefore, the electric energy stored in the
cloud is just the electric energy density times
the volume of the cloud.
Example – A Thunderstorm
Diagram
r = 10 km
Thundercloud
volume = π r2 h
h = 10 km
Example – A Thunderstorm
Calculation
The electric energy density in the cloud is
uE = ½ ε0 E2 = 4.4 x 10-2 J/m3.
The volume of the cloud is, v = π r2 h, that is,
v = 3.1 x 1012 m3.
Therefore, the total electric energy stored in
the cloud is U = uEv = 140 GJ.
Using Capacitors
The Effect of Dielectrics
Michael Faraday discovered that the capacitance
increases when the space between conductors is
replaced by a dielectric.
Today, we understand this to
be a consequence of the
polarization of molecules.
Michael Faraday
1791 – 1867
wikimedia
The Effect of Dielectrics
−σb
The polarized molecules
of the dielectric tend to
align themselves parallel
to the electric field created
by the charges on the
conductors
+
+
+
+
+
+
+
+
+σb
-
The Effect of Dielectrics
The bound charge σb
induced on the surface
of the dielectric creates
an electric field
opposed to the electric
field of the free charge
σf on the conductors,
thereby reducing the
field between them.
Cuál es el mecanismo?
The Effect of Dielectrics
The reduction in electric field strength from the initial
field E0 to the reduced field E is quantified by the
dielectric constant κ (kappa)
E=
E0
κ
The dielectric increases the capacitance
by the same factor κ.
The Effect of Dielectrics
For a parallel plate capacitor, with a dielectric
between the plates, the electric field is
E = Q /(κε 0 A)
The product of the dielectric constant κ and the
permittivity of free space ε0
is called the
ε = κε 0 permittivity
Capacitores en paralelo
At equilibrium, the potential across each capacitor
is the same, namely, 12 V
same potential
same potential
Capacitores en paralelo
The two
capacitors
are equivalent
to a single
capacitor with
capacitance
C= C1 + C2
Capacitores en paralelo


The flow of charges ceases when the voltage across the
capacitors equals that of the battery
The potential difference across the capacitors is the same




And each is equal to the voltage of the battery
∆V1 = ∆V2 = ∆V
∆V is the battery terminal voltage
The capacitors reach their maximum charge when the
flow of charge ceases
The total charge is equal to the sum of the charges on the
capacitors

Qtotal = Q1 + Q2
Capacitores en paralelo

The
capacitors
can
be
replaced with one capacitor
with a capacitance of Ceq

The
equivalent
capacitor
must have exactly the same
external effect on the circuit
as the original capacitors
Capacitores en paralelo



Ceq = C1 + C2 + C3 + …
The equivalent capacitance of a parallel
combination of capacitors is greater than any of the
individual capacitors
 Essentially, the areas are combined
Use the active figure to vary the battery potential
and the various capacitors and observe the
resulting charges and voltages on the capacitors
Capacitores en Serie
The sum of the
potentials across
both capacitors
will be equal to
12 V
Capacitores en serie
The potential V1
across C1 plus the
potential V2 across
C2 is equal to the
potential difference V
between points a and b:
V = V1 + V2
=
1/
C 1/ C1 + 1/ C2
Capacitores en serie

When a battery is
connected to the circuit,
electrons are transferred
from the left plate of C1 to
the right plate of C2
through the battery
Capacitores en serie


As this negative charge accumulates on the right
plate of C2, an equivalent amount of negative charge
is removed from the left plate of C2, leaving it with
an excess positive charge
All of the right plates gain charges of –Q and all the
left plates have charges of +Q
Capacitores en serie


An equivalent capacitor
can be found that
performs the same
function as the series
combination
The charges are all the
same
Q1 = Q 2 = Q
Capacitores en serie


The potential differences add up to the battery voltage
ΔVtot = ∆V1 + ∆V2 + …
The equivalent capacitance is
1
1
1
1
=
+
+
+
Ceq C1 C2 C3

The equivalent capacitance of a series combination is
always less than any individual capacitor in the
combination
Energía almacenada en un Capacitor


Assume the capacitor is being charged and, at
some point, has a charge q on it
The work needed to transfer a charge from one
plate to the other is
q
dW =
∆Vdq =
dq
C

The total work required is
q
Q2
=
W ∫=
dq
0 C
2C
Q
Energy, cont
The work done in charging the capacitor appears
as electric
potential energy U:
2
Q
1
1
U=
=
Q ∆V =
C ( ∆V ) 2
2C 2
2




This applies to a capacitor of any geometry
The energy stored increases as the charge
increases and as the potential difference
increases
In practice, there is a maximum voltage before
discharge occurs between the plates
Energy, final



The energy can be considered to be stored in
the electric field
For a parallel-plate capacitor, the energy can
be expressed in terms of the field as U = ½
(εoAd)E2
It can also be expressed in terms of the
energy density (energy per unit volume)
uE = ½ εoE2
Resumen





Capacitancia
 Placas paralelas
Capacitores
 En paralelo
 En serie
Energia almacenada
Densidad de Energía
Efecto del dielectrico
C = Q / V (faradios)
C = ε0 A/d
C = C1 + C2
1/C = 1/C1 + 1/C2
U = ½ QV
uE = ½ ε0 E2
E = E0 / κ
TT: Efecto Piezoelectrico
https://www.youtube.com/wat
ch?v=uNKMD8Z3-e4
Some Uses of Capacitors

Defibrillators



When cardiac fibrillation occurs, the heart produces a
rapid, irregular pattern of beats
A fast discharge of electrical energy through the heart
can return the organ to its normal beat pattern
In general, capacitors act as energy reservoirs
that can be slowly charged and then discharged
quickly to provide large amounts of energy in a
short pulse
Dielectrics – An Atomic View


The molecules that make
up the dielectric are
modeled as dipoles
The molecules are
randomly oriented in the
absence of an electric
field
Dielectrics – An Atomic View, 2



An external electric
field is applied
This produces a torque
on the molecules
The molecules partially
align with the electric
field
Dielectrics – An Atomic View,
final


An external field can
polarize the dielectric
whether the molecules are
polar or nonpolar
The charged edges of the
dielectric act as a second pair
of plates producing an
induced electric field in the
direction opposite the
original electric field
Dielectrics – An Atomic View,
3


The degree of alignment of the molecules
with the field depends on temperature and
the magnitude of the field
In general,
 the alignment increases with decreasing
temperature
 the alignment increases with increasing
field strength
Capacitors with Dielectrics

A dielectric is a nonconducting material that, when
placed between the plates of a capacitor, increases the
capacitance


Dielectrics include rubber, glass, and waxed paper
With a dielectric, the capacitance becomes
κCo


C=
The capacitance increases by the factor κ when the dielectric
completely fills the region between the plates
κ is the dielectric constant of the material
Dielectrics, cont



For a parallel-plate capacitor, C = κεo(A/d)
In theory, d could be made very small to create a
very large capacitance
In practice, there is a limit to d


d is limited by the electric discharge that could occur
though the dielectric medium separating the plates
For a given d, the maximum voltage that can be
applied to a capacitor without causing a
discharge depends on the dielectric strength of
the material
Dielectrics, final

Dielectrics provide the following advantages:
 Increase in capacitance
 Increase the maximum operating voltage
 Possible mechanical support between the
plates
• This allows the plates to be close
together without touching
• This decreases d and increases C
Types of Capacitors – Tubular


Metallic foil may be
interlaced with thin
sheets of paraffinimpregnated paper or
Mylar
The layers are rolled
into a cylinder to form a
small package for the
capacitor
Types of Capacitors – Oil Filled


Common for highvoltage capacitors
A number of interwoven
metallic plates are
immersed in silicon oil
Types of Capacitors – Electrolytic


Used to store large
amounts of charge at
relatively low voltages
The electrolyte is a
solution that conducts
electricity by virtue of
motion of ions
contained in the solution
Types of Capacitors – Variable




Variable capacitors consist
of two interwoven sets of
metallic plates
One plate is fixed and the
other is movable
These capacitors generally
vary between 10 and 500 pF
Used in radio tuning circuits
Electric Dipole



An electric dipole consists of
two charges of equal
magnitude and opposite
signs
The charges are separated by
2a

The electric
p dipole moment
( ) is directed along the line
joining the charges from –q
to +q
Energy Stored in a Capacitor
Thus, the total amount of work required to charge the capacitor from q = 0
to a final charge of q = Q is
Q
1 Q
Q2
q
W = ∫ dq = ∫ qdq =
C
C 0
2C
0
But, in an isolated system with no non-conservative forces, total mechanical
energy must be conserved.
Therefore, the work done to charge the capacitor must equal the change in
the system’s potential energy.
Energy Stored in a Capacitor
Q2 1
1
2
= Q∆V = C (∆V )
U=
2C 2
2
Energy stored in a charged capacitor
Energy Stored in a Capacitor
Q2 1
1
2
U=
= Q∆V = C (∆V )
2C 2
2
-It’s not obvious, but the potential energy stored in the capacitor actually resides in
its electric field.
-This implies we should be able to solve the density of the energy stored in the field
(J/m3).
-For a parallel plate capacitor, we already know: ∆V = Ed
ε A
-and, its capacitance is just: C = 0
d
-Substituting these into the purple equation,
1 ε0 A 2 2 1
(
U=
E d ) = (ε 0 Ad )E 2
2 d
2
-Dividing by the volume in between the plates of the capacitor (V=Ad), we
get
1
uE = ε 0 E 2
2
Energy per unit volume in
a capacitor (J/m3)
Energy Stored in a Capacitor
-We don’t attempt it here, but it can be shown that this result is valid for any electric
field!
1
uE = ε 0 E 2
2
Energy per unit volume in
an electric field.
-In a very real sense, electric fields “carry” energy.
Rewiring two Charged Capacitors
Two capacitors, C1 and C2 (C1 > C2), are charged to the same initial potential difference, ΔVi.
The charged capacitors are removed from the battery, and their plates are connected with
opposite polarity, as shown. The switches, S1 and S2, are then closed.
(a)
(b)
Find the final potential difference ΔVf between a and b after the switches are closed.
Find the total energy stored in the capacitors before and after the switches are closed and
determine the ratio of the final energy to the initial energy.
What happens?
Capacitors with dielectrics
Consider parallel-plate capacitor where ΔV0 = Q0/C0
Assume no battery is connected  Q can’t change
When you stick a dielectric in between the plates 
-where κ is a dimensionless
constant called the
“dielectric constant”
∆V0
∆V =
κ
Capacitors with dielectrics
-Q on the capacitor does not change
-Therefore:
C=
Q0
Q0
=
∆V ∆V0
C = κC0
-the capacitance is changed
by a factor of κ.
-as κ goes up, C goes up.
=κ
κ
Q0
∆V0
Capacitors with dielectrics
-For a parallel plate capacitor
C0 =
ε0 A
d
⇒C =κ
ε0 A
d
To make capacitance ↑
-decrease d
-increase A
-increase κ
- Only limited by
“dielectric strength” of the
dielectric
Example values of dielectric constant
“Dielectric strength” is the
maximum field in the
dielectric before breakdown.
(a spark or flow of charge)
E max = Vmax / d
EG 26.5 – Energy stored before and after
A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is
then removed, and a slab of material that has a dielectric constant κ is inserted
between the plates. Identify the system as the capacitor and the dielectric.
Find the energy stored in the system before and after the dielectric is inserted.
Rewiring two Charged Capacitors
A parallel-plate capacitor is charged with a battery to a charge of Q0. The battery is
then removed, and a slab of material that has a dielectric constant κ is inserted
between the plates. Identify the system as the capacitor and the dielectric.
Find the energy stored in the system before and after the dielectric is inserted.
Before:
Q02
U0 =
2C0
After:
Q02
U0 =
2C
Q02
U
U0 =
= 0
2κC0 κ
Where did the energy go?
Electric Dipole in an Electric Field
The combination of two equal charges of opposite sign, +q and –q,
separated by a distance 2a
Every dipole can be characterized by it’s “dipole moment.”
- vector which points from –q to +q
-magnitude p = 2aq

p1

p2
  
p= p1 + p 2
Electric Dipole in an Electric Field
What happens when we pop this baby in an external E-field?
Electric Dipole in an Electric Field
What happens when we pop this baby in an external E-field?
-external field exerts F=qE on each charge
-net torque about the dipole’s center
-dipole rotates to “align” with the field
Electric Dipole in an Electric Field
What happens when we pop this baby in an external E-field?
-external field exerts F=qE on each charge
-net torque about the dipole’s center
-dipole rotates to “align” with the field
τ + = Fa sin θ
τ − = Fa sin θ
Electric Dipole in an Electric Field
τ net = 2Fa sin θ
but,
F = qE and p = 2aq
Thus:
τ = 2aqE sin θ = pE sin θ
Electric Dipole in an Electric Field
τ net = 2Fa sin θ
but,
F = qE and p = 2aq
Thus:
τ = 2aqE sin θ = pE sin θ
 
τ = p× E

Electric Dipole in an Electric Field
The dipole and the external field are a system
-electric force is an internal conservative force
we can describe its work using a potential
energy
In other words, different configurations of the
dipole-field system have different potential
energies.
Electric Dipole in an Electric Field
As the dipole aligns with the field, the
system’s potential energy goes down.
Electric Dipole in an Electric Field
-Work must be done to “un-align” the dipole
from the field.
-in an isolated system, the work input must
correspond to an increase in potential energy.
Electric Dipole in an Electric Field
-Work must be done to “un-align” the dipole
from the field.
-in an isolated system, the work input must
correspond to an increase in potential energy.
W = ΔK + ΔU
Electric Dipole in an Electric Field
-To rotate the dipole through some small angle dθ, an amount dW of work must
be done.
dW = τdθ
but,
τ = pE sin θ
Electric Dipole in an Electric Field
-To rotate the dipole through some small angle dθ, an amount dW of work must
be done.
dW = τdθ
but,
τ = pE sin θ
-so, to rotate the dipole from θi to θf, the change in potential energy is:
θf
θf
θf
θi
θi
θi
U f − U i = ∫ τdθ = ∫ pE sin θdθ = pE ∫ sin θdθ
θ
= pE[− cos θ ]θif = pE (cos θ i − cos θ f )
Electric Dipole in an Electric Field
Let’s define the zero potential energy as being when the dipole is at θ = 90,
Ui = 0
when
θ i = 90
Electric Dipole in an Electric Field
Let’s define the zero potential energy as being when the dipole is at θ = 90,
Ui = 0
when
θ i = 90
We’ll use this reference energy as an anchor point.
At any time, we can write the system’s
instantaneous potential energy, U, with respect to
the zero-point potential energy.
U = U f − U i = U f − U θ =90 = U f − 0
Electric Dipole in an Electric Field
Let’s define the zero potential energy as being when the dipole is at θ = 90,
Ui = 0
when
θ i = 90
We’ll use this reference energy as an anchor point.
At any time, we can write the system’s
instantaneous potential energy, U, with respect to
the zero-point potential energy.
U = U f − U i = U f − U θ =90 = U f − 0
But, we already know
θ
= pE[− cos θ ]θif = pE (cos θ i − cos θ f ) = − pE cos θ f
U = − pE cos θ
Electric Dipole in an Electric Field
Let’s define the zero potential energy as being when the dipole is at θ = 90,
Ui = 0
when
θ i = 90
We’ll use this reference energy as an anchor point.
At any time, we can write the system’s
instantaneous potential energy, U, with respect to
the zero-point potential energy.
U = U f − U i = U f − U θ =90 = U f − 0
But, we already know
θ
= pE[− cos θ ]θif = pE (cos θ i − cos θ f ) = − pE cos θ f
U = − pE cos θ
 
U = −p•E
The Water Molecule
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample
contains 1021 water molecules. All of the dipoles are oriented in the direction
of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to
one in all the dipoles are perpendicular to the external field (θ = 90)?
The Water Molecule
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample
contains 1021 water molecules. All of the dipoles are oriented in the direction
of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to
one in all the dipoles are perpendicular to the external field (θ = 90)?
∆U = W
W = U 90° − U 0° = (− NpE cos 90) − (− NpE cos 0)
= NpE = (10 21 )(6.3 ×10 −30 C ⋅ m)(2.5 ×105 N / C )
= 1.6 ×10 −3 J
The Water Molecule
A water molecule has an electric dipole moment of 6.3x10-30 Cm. A sample
contains 1021 water molecules. All of the dipoles are oriented in the direction
of an external E-field, which has a magnitude of 2.5x105 N/C.
How much work is required to rotate all the dipoles from this orientation (θ = 0) to
one in all the dipoles are perpendicular to the external field (θ = 90)?
∆U = W
W = U 90° − U 0° = (− NpE cos 90) − (− NpE cos 0)
= NpE = (10 21 )(6.3 ×10 −30 C ⋅ m)(2.5 ×105 N / C )
= 1.6 ×10 −3 J
Example P26.9
When a potential difference of 150 V is applied to the plates
of a parallel-plate capacitor, the plates carry a surface charge
density of 30.0 nC/cm2. What is the spacing between the
plates?
∈0 A
=
Q
( ∆V )
d
∈0 ( ∆V )
=
d =
σ
( 8.85 × 10
( 30.0 × 10
−9
)
C2 N ⋅ m 2 ( 150 V )
=
C cm 2 1.00 × 104 cm 2 m 2
−12
)(
)
4.42 µm
Example P26.21
Four capacitors are connected as shown in
Figure P26.21.
(a) Find the equivalent capacitance between
points a and b.
(b) Calculate the charge on each capacitor if
ΔVab = 15.0 V.
1
1
1
=
+
Cs 15.0 3.00
Q = C∆V = ( 5.96 µF) ( 15.0 V ) = 89.5 µ C
Q 89.5 µ C
=
= 4.47 V
C 20.0 µF
15.0 − 4.47 =
10.53 V
∆V =
Cs = 2.50 µF
Cp = 2.50 + 6.00 = 8.50 µF

1
1 
Ceq =
 8.50 µF + 20.0 µF 
−1
=5.96 µF
Q = C∆V = ( 6.00 µF) ( 10.53 V ) = 63.2 µ C on 6.00 µF
89.5 − 63.2 =
26.3 µ C
Example P26.27
Find the equivalent capacitance between
points a and b for the group of capacitors
connected as shown in Figure P26.27. Take
C1 = 5.00 μF, C2 = 10.0 μF, and C3 = 2.00
μF.
1 
 1
+
Cs =


5.00 10.0 
−1
3.33 µF
=
= 8.66 µF
2( 3.33) + 2.00
Cp=
1
=
Cp2 2=
( 10.0) 20.0 µF
1 
 1
+
Ceq =


8.66 20.0 
−1
=6.04 µF
Example P26.35
A parallel-plate capacitor is charged and then disconnected
from a battery. By what fraction does the stored energy
change (increase or decrease) when the plate separation is
doubled?
d2 = 2d1
,
C2 =
1
C1
2
. Therefore, the
stored energy doubles
Example P26.43
Determine (a) the capacitance and (b) the
maximum potential difference that can be applied
to a Teflon-filled parallel-plate capacitor having a
plate area of 1.75 cm2 and plate separation of 0.040
0 mm.
κ ∈0 A
C=
d
(
)(
)
2.10 8.85 × 10−12 F m 1.75 × 10−4 m 2
=
=
8.13 × 10−11 F =
81.3 pF
−5
4.00 × 10 m
(
)(
)
∆Vmax = Emax d = 60.0 × 106 V m 4.00 × 10−5 m = 2.40 kV
Example P26.59
A parallel-plate capacitor is constructed using a dielectric
material whose dielectric constant is 3.00 and whose
dielectric strength is 2.00 × 108 V/m. The desired
capacitance is 0.250 μF, and the capacitor must withstand a
maximum potential difference of 4 000 V. Find the minimum
area of the capacitor plates.
κ = 3.00
=
C
=
A
∆Vmax
Emax =×
2.00 108 V m =
d
κ ∈0 A
= 0.250 × 10−6 F
d
(
(
)
)(
0.250 × 10−6 ( 4 000)
C∆Vmax
Cd
=
=
=
κ ∈0 κ ∈0 Emax 3.00 8.85 × 10−12 2.00 × 108
)
0.188 m 2
Electric Circuits

A circuit is a collection of objects usually containing a
source of electrical energy connected to elements that
convert electrical energy to other forms

A circuit diagram – a simplified representation of an
actual circuit – is used to show the path of the real circuit

Circuit symbols are used to represent various elements